# Fractional-Linear Inequalities

Here, we will consider inequalities of the form (ax+b)/(cx+d)>0 or (ax+b)/(cx+d)<0, where a!=0,c!=0.

Example. Solve inequality (3x+7)/(2x-7)>5.

In cases where right-hand side doesn't equal 0, we need to transform inequality into equivalent one, where right-hand side equals 0:

(3x+7)/(2x-7)>5;

(3x+7)/(2x-7)-5>0;

(3x+7)/(2x-7)-(5(2x-7))/(2x-7)>0;

(3x+7-5(2x-7))/(2x-7)>0;

(3x+7-10x+35)/(2x-7)>0;

(-7x+42)/(2x-7)>0; divide both sides by -7:

(x-6)/(2x-7)<0;

Fraction is negative in two cases:

1. Numerator is negative, denominator is positive: {(x-6<0),(2x-7>0):}.
2. Numerator is positive, denominator is negative: {(x-6>0),(2x-7<0):}.

Therefore, we have set of two systems (recall that set contains solutions that satisfy at least one system).

Set of solutions of the first system is x<6 and x>3.5, i.e. 3.5<x<6.

Set of solutions of the second system is x>6 and x<3.5, i.e. there are no solutions.

Therefore, set of solutions of initial inequality is interval (3.5,6).